Finding FE, Tension & Acceleration in a 3-Block Frictionless Ramp Setup

[watever]

three blocks on a frictionless 45° ramp held together by massless ropes
top block's mass, m1=1.00 kg.
middle block's mass, m2=2.00 kg
bottom block's mass, m3=3.00 kg.
1. the force 'FE' is exerted on block one and is just enough to hold all the blocks in place
find FE and the tensions of the ropes
2. FE is no longer acting on the blocks so they slide down the ramp, find the acceleration of the blocks and the tensions in the ropes
3. 2FE is now acting on block one, find acceleration of blocks and tensions in the ropes


All I really need help with is the tension

 

[blochwave]

Basically it's the force coming from the rope

So for example, the top block is sitting still for part 1 and has what amounts to 5kg of mass hanging on that segment of rope. Figure out the force of gravity going down the incline caused by the 5kg, that force will be the tension in the rope, it's how much force the rope has to apply to hold them there

The tension in the other segment(connecting blocks 2 and 3), well that segment of rope is only supporting the 3 kg mass

 

[Moetasim]

I would approach this problem by first drawing a free-body diagram for each block and analyzing the forces acting on them. Since the ramp is frictionless, the only forces acting on the blocks are the gravitational force (mg) and the tension forces from the ropes.

1. To find the force FE, we can use the fact that it is just enough to hold all the blocks in place. This means that the net force on the blocks is zero, so the tension forces in the ropes must balance out the gravitational force. Using this information, we can set up the following equations:

For block 1:
FE = T1sin45°

For block 2:
T1sin45° = T2sin45° + m2g

For block 3:
T2sin45° = T3sin45° + m3g

Solving these equations simultaneously, we can find the value of FE and the tensions in the ropes.

2. When FE is no longer acting on the blocks and they start sliding down the ramp, the net force on each block will be equal to their respective mass times the acceleration (F = ma). We can set up the following equations:

For block 1:
T1sin45° = m1a

For block 2:
T2sin45° = m2a

For block 3:
T3sin45° = m3a

Solving these equations simultaneously, we can find the acceleration of the blocks and the tensions in the ropes.

3. When 2FE is now acting on block one, the net force on block one will be equal to its mass times the acceleration, and the net force on blocks 2 and 3 will be equal to their respective mass times the acceleration plus the tension force from the ropes. We can set up the following equations:

For block 1:
2FE - T1sin45° = m1a

For block 2:
T1sin45° - T2sin45° = m2a

For block 3:
T2sin45° - T3sin45° = m3a

Solving these equations simultaneously, we can find the acceleration of the blocks and the tensions in the ropes.

In conclusion, by carefully analyzing the forces acting on the blocks and using Newton's laws of motion, we can find the tension forces in the ropes and the acceleration of the blocks in each scenario. This approach can be